Algebra and Functions (Edexcel IAL P2)

Edexcel International A-Level Mathematics· 2018 specification (Issue 3), P2 §2.1· 20 min read

1. Algebraic Division by Linear Divisors★★☆☆☆⏱ 5 min

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📘 Definition

Polynomial Division Rules

When a degree $n$ polynomial (dividend) is divided by a linear (degree 1) divisor, the result is a degree $n - 1$ quotient polynomial plus a constant remainder.

Example:

Dividing $x^{3} + 3 x^{2} - 4$ by $(x + 2)$ gives quotient $x^{2} + x - 2$ and remainder 0.

For P2, you only need to divide by linear divisors of the form $(a x \pm b)$ . You may use either long division or coefficient matching (equating coefficients method) for this process, both are accepted by Edexcel examiners as long as full working is shown.

📐 Worked Example

Find the quotient and remainder when $f (x) = 2 x^{3} + 5 x^{2} - x - 6$ is divided by $(x + 2)$ .

1
Use coefficient matching: let $2 x^{3} + 5 x^{2} - x - 6 = (x + 2) (A x^{2} + B x + C) + R$
2
Expand the right-hand side: $A x^{3} + (2 A + B) x^{2} + (2 B + C) x + (2 C + R)$
3
Equate coefficients of matching powers of $x$ :
4
$A = 2 \text{ (}x^3 \text{ term)}$
5
$2A + B = 5 \implies 4 + B =5 \implies B=1 \text{ (}x^2 \text{ term)}$
6
$2B + C = -1 \implies 2 + C = -1 \implies C=-3 \text{ (}x \text{ term)}$
7
$2C + R = -6 \implies -6 + R = -6 \implies R=0 \text{ (constant term)}$
8
State final result: Quotient = $2 x^{2} + x - 3$ , Remainder = 0

Exam tip:

Always label quotient and remainder explicitly, as exam mark schemes allocate separate marks for each term.

2. Remainder Theorem★★☆☆☆⏱ 5 min

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📘 Definition

Remainder Theorem

$f (\frac{b}{a})$

For any polynomial $f (x)$ , the remainder when $f (x)$ is divided by $(a x - b)$ is equal to $f (\frac{b}{a})$ . For divisor $(a x + b)$ , the remainder is $f (- \frac{b}{a})$ .

📐 Worked Example

Find the remainder when $f (x) = 6 x^{3} + 11 x^{2} - x - 6$ is divided by $(2 x - 1)$ .

1
For divisor $(2 x - 1)$ , $a = 2$ , $b = 1$ , so evaluate $f (\frac{1}{2})$
2
$f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^3 + 11\left(\frac{1}{2}\right)^2 - \frac{1}{2} - 6$
3
$= 6\left(\frac{1}{8}\right) + 11\left(\frac{1}{4}\right) - \frac{1}{2} -6$
4
$= \frac{3}{4} + \frac{11}{4} - \frac{2}{4} - 6 = \frac{12}{4} -6 = 3 -6 = -3$
5
Final remainder = $- 3$

3. Factor Theorem & Cubic Factorisation★★★☆☆⏱ 6 min

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📘 Definition

Factor Theorem

If the remainder when $f (x)$ is divided by $(a x - b)$ is 0, then $(a x - b)$ is a factor of $f (x)$ . In other words, if $f (\frac{b}{a}) = 0$ , $(a x - b)$ is a factor.

To factorise a cubic polynomial: first test small integer values to find a root, use the Factor Theorem to get a linear factor, then use coefficient matching or division to find the quadratic factor, then factorise the quadratic if possible.

📐 Worked Example

Factorise fully $f (x) = x^{3} + 3 x^{2} - 4$ .

1
Test small integer values: $f (1) = 1 + 3 - 4 = 0$ , so $(x - 1)$ is a factor
2
Let $x^{3} + 3 x^{2} - 4 = (x - 1) (A x^{2} + B x + C)$
3
Expand RHS and equate coefficients: $A = 1$ , $B = 4$ , $C = 4$
4
Quadratic factor is $x^{2} + 4 x + 4$ , which is a perfect square: $(x + 2)^{2}$
5
Fully factorised form: $(x - 1) (x + 2)^{2}$

Exam tip:

Only test integer values of $x$ that divide the cubic's constant term first, as this reduces the number of tests needed.

4. Combined Exam-Style Problems★★★★☆⏱ 4 min

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Most P2 exam questions combine algebraic division, Remainder Theorem and Factor Theorem into multi-part 4-6 mark questions, often requiring you to solve for unknown coefficients first.

📐 Worked Example

Given $f (x) = 2 x^{3} + a x^{2} + b x - 6$ . When $f (x)$ is divided by $(x - 1)$ the remainder is 0; when divided by $(x + 2)$ the remainder is 6. (a) Find $a$ and $b$ . (b) Factorise $f (x)$ fully.

1
Part (a): Remainder Theorem for $(x - 1)$ : $f (1) = 2 + a + b - 6 = 0 ⟹ a + b = 4$ .
2
Remainder Theorem for $(x + 2)$ : $f (- 2) = - 16 + 4 a - 2 b - 6 = 6 ⟹ 4 a - 2 b = 28 ⟹ 2 a - b = 14$ .
3
Solve simultaneously ( $a + b = 4$ , $2 a - b = 14$ ): adding gives $3 a = 18$ , so $a = 6$ , $b = - 2$ .
4
Part (b): $f (x) = 2 x^{3} + 6 x^{2} - 2 x - 6 = 2 (x^{3} + 3 x^{2} - x - 3)$ .
5
Factor by grouping: $x^{3} + 3 x^{2} - x - 3 = x^{2} (x + 3) - (x + 3) = (x + 3) (x^{2} - 1) = (x + 3) (x - 1) (x + 1)$ .
6
Final factorised form: $f (x) = 2 (x - 1) (x + 1) (x + 3)$ .

✓ Quick check

Independent practice: For $f (x) = 3 x^{3} - 5 x^{2} - 4 x + 4$ , find the remainder when divided by $(3 x + 1)$ , then fully factorise $f (x)$ . Show all working.

5. Common Pitfalls

Wrong move:

Using $x = \frac{b}{a}$ for divisor $(a x + b)$ instead of $x = - \frac{b}{a}$

Why:

Incorrect sign substitution leads to wrong remainders or missed linear factors

Correct move:

Rewrite $(a x + b)$ as $(a x - (- b))$ first, so use $x = - \frac{b}{a}$ for substitution

Wrong move:

Failing to label quotient and remainder separately when asked

Why:

Exam mark schemes allocate separate marks for each term, so unlabeled answers lose points

Correct move:

Explicitly write 'Quotient = [expression]' and 'Remainder = [value]' in your answer

Wrong move:

Testing random integer values for the Factor Theorem

Why:

Wastes exam time and increases risk of arithmetic error

Correct move:

Only test integer values of $x$ that divide the constant term of the cubic (e.g. for constant -6, test ±1, ±2, ±3, ±6)

Wrong move:

Leaving a factorisable quadratic factor unfactorised

Why:

Questions ask for full factorisation, so partial answers lose marks

Correct move:

Always check if the quadratic factor can be split into linear terms or is a perfect square

Wrong move:

Using full long division when coefficient matching is faster

Why:

Long division has higher risk of sign errors for linear divisors

Correct move:

Use coefficient matching for linear divisors to save time and reduce error

6. Quick Reference Cheatsheet

Concept	Rule / Formula	Exam Use Case
Algebraic Division (linear divisor)	$f (x) = (a x \pm b) Q (x) + R$ , $de g (Q) = de g (f) - 1$ , $R$ constant	Find quotient and remainder of polynomial division
Remainder Theorem	Remainder = $f (\frac{b}{a})$ for divisor $(a x - b)$	Calculate remainder without full algebraic division
Factor Theorem	If $f (\frac{b}{a}) = 0$ , $(a x - b)$ is a factor of $f (x)$	Find linear factors of cubic polynomials
Cubic Factorisation Steps	Test root → 2. Get linear factor → 3. Find quadratic factor → 4. Factorise quadratic	Fully factorise cubic expressions

7. Frequently Asked

Do I need to learn division by quadratic divisors for P2?

No, Edexcel IAL P2 only requires division of polynomials by linear divisors of the form $(a x + b)$ or $(a x - b)$ . Quadratic divisors are not assessed in this unit.

Can I use a calculator for algebraic division questions?

Calculators are permitted for all P2 papers, but you must show full written working for algebraic division and factorisation steps to earn full marks, even if you check answers with your calculator.

Going deeper

syllabusEdexcel IAL Maths P2 Official SpecificationFull specification for the WMA12 P2 unit

What's Next

Now that you have mastered algebraic division, Factor and Remainder Theorems for Edexcel IAL P2, you are ready to progress to other core P2 topics. These foundational algebra skills are required for coordinate geometry, differentiation, and trigonometry topics later in P2, as well as for advanced polynomial work in P3 and P4. These concepts are tested in almost every P2 paper, often combined with other topics in multi-part questions, so practice regularly to build speed and accuracy. Always show full working for all steps, even if you use a calculator to check your answers, to ensure you earn all available method marks in your exam.